Question

Solve sin^2 x-5 sinx+6=0 for all solutions [0,2pi)?

Answer 1

No real solutions.

Explanation:

`sin^2(x)-5sin(x)+6 = 0`

`=> (sin(x)-2)(sin(x)-3) = 0`

`=> sin(x)-2 = 0 or sin(x)-3 = 0`

`=> sin(x) = 2 or sin(x) = 3`

However, `sin(x) in [-1, 1]` for all `x in RR`, meaning neither of the above equations has a solution in `[0, 2pi)`. Indeed, the given equation has no real solutions.